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Ex.9.2 Q1 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

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Question

In the given figure, \(ABCD\) is parallelogram, \(AE ⊥ DC\) and \(CF ⊥ AD\) . If \(AB = 16\; \rm{cm}\), \(AE = 8 \;\rm{cm}\) and \(CF = 10\;\rm{cm}\), find \(AD\).

 

 Video Solution
Areas Of Parallelograms And Triangles
Ex 9.2 | Question 1

Text Solution

What is known?

ABCD is parallelogram \({AE}\bot {DC}\) , and  \({CF}\bot {AD}\) . If \(AB = 16 \,\rm{cm}, AE = 8 \,\rm{cm}\) and \(CF = 10\, \rm{cm}\), find \(AD\).

What is unknown?

Length of \(AD\).

Reasoning:

Area of a parallelogram \(ABCD\)

\[\begin{align}&= CD \!\!\times\!\!{ AE}\,\\&=\,\,{AD}\,\!\!\times\!\!\,{CF}\end{align}\]

Steps:

Given:

\(AB = 16\; \rm{cm}\).

\(AE = 8 \;\rm{cm}\).

\(CF = 10\;\rm{cm}\).

In parallelogram \(ABCD\), \(CD = AB = 16 \;\rm{cm}\)

[Opposite sides of a parallelogram are equal]

We know that,

Area of a parallelogram \(=\) Base \(×\) Corresponding altitude

Therefore, Area of a parallelogram \(\begin{align} {\rm{ABCD =  CD \times AE}}\,{\rm{ = }}\,\,{\rm{AD}}\,\,{\rm{ \times }}\,\,{\rm{CF}}\end{align} \)

Here \(CD\) and \(AD\) act as base and \(AE ⊥ CD\) , \(CF ⊥ AD\) are the corresponding altitudes.

Therefore,\[\begin{align} {AD} &= \frac{{{{CD }} \times {{ AE}}}}{{{{CF}}}}\end{align} \]

Substituting the values for \(CD\) , \(AE\) and \(CF\), we get

\[\begin{align}{AD}&=\frac{16 \times 8}{10} {cm}=12.8 {cm}\end{align}\]

Thus, the length of \(AD\) is \(12.8 \;\rm{cm}\).

 Video Solution
Areas Of Parallelograms And Triangles
Ex 9.2 | Question 1
  
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